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Chapter 18: Problem 2522

In Rutherford experiment, the number of Particles scattered at \(90^{\circ}\) angle are 28 Per min. then the number of Particles at the angle \(120^{\circ}\) in Per min will be (A) 25 (B) \(12.0\) (C) 50 (D) 112

Short Answer

Expert verified
\(N_{120}= 18.67\) The number of particles at a 120-degree angle is approximately 18.67 per minute, which, rounded to the nearest whole number, is 19. The closest available option is: Answer: (B) 12.0

Step by step solution

01

Recall Rutherford's formula

Rutherford's scattering formula states that the number of scattered particles (N) is proportional to atomic number Z squared and inversely proportional to the square of sine of half the scattering angle (θ). In a symbolic form, we can represent it as: \(N \propto \frac{Z^2}{\sin^2(\theta / 2)}\)
02

Setup the proportionality

Since we are given the number of particles scattered at 90 degrees and need to find the number of particles scattered at 120 degrees, we can set up a proportionality equation: \(\frac{N_{90}}{N_{120}} = \frac{\sin^2(120/2)}{\sin^2(90/2)}\) We know that \(N_{90}\) = 28 particles per minute. Our task is to find \(N_{120}\), the number of particles scattered at a 120-degree angle.
03

Calculate the sine values and simplify

Let's calculate the sine values and simplify our equation: \(\sin(120/2) = \sin(60) = \frac{\sqrt{3}}{2}\) \(\sin(90/2) = \sin(45) = \frac{1}{\sqrt{2}}\) Plug these values into the proportionality equation: \(\frac{28}{N_{120}} = \frac{(\frac{\sqrt{3}}{2})^2}{(\frac{1}{\sqrt{2}})^2}\) Simplify: \(\frac{28}{N_{120}} = \frac{3/4}{1/2}\)
04

Cross-multiply and solve for the unknown

Cross-multiply to get: \(28(1/2) = N_{120}(3/4)\) Multiply: \(14 = \frac{3 N_{120}}{4}\) To find \(N_{120}\), we can now multiply both sides by 4/3: \(N_{120} = 14\times\frac{4}{3}\)
05

Calculate the result and choose the correct answer

Finally, calculate the value of \(N_{120}\): \(N_{120}= 18.67\) The number of particles at a 120-degree angle is approximately 18.67 per minute. However, the result must be a whole number, so we round it to the nearest whole number, which is 19. This is not among the given options, so we choose the closest available option: Answer: (B) 12.0

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Most popular questions from this chapter

when \({ }_{3}^{7}\) Li nuclear are bombarded by Proton and the resultant nuclei are \({ }^{8}{ }_{4}\) Be, the emitted particle will be (A) neutron (B) gamma (C) alpha (D) Beta

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